ARCHIVUM MATHEMATICUM (BRNO)
Tomus 45 (2009), 1–18
ON OSCILLATION CRITERIA FOR THIRD ORDER
NONLINEAR DELAY DIFFERENTIAL EQUATIONS
Ravi P. Agarwal, Mustafa F. Aktas, and A. Tiryaki
Abstract. In this paper we are concerned with the oscillation of third order
nonlinear delay differential equations of the form
r2 (t) r1 (t) x0
0 0
+ p (t) x0 + q (t) f (x (g (t))) = 0.
We establish some new sufficient conditions which insure that every solution
of this equation either oscillates or converges to zero.
1. Introduction
In this paper we consider nonlinear third order functional differential equations
of the form
0 0
(1.1)
r2 (t) (r1 (t) x0 ) + p (t) x0 + q (t) f (x (g (t))) = 0 ,
where r1 , r2 , p, q ∈ C (I, R), I = [t0 , ∞) ⊂ R, t0 ≥ 0 is a constant such that r1 > 0,
r2 > 0, p(t) ≥ 0, q (t) ≥ 0, q (t) 6≡ 0 in the neighborhood of ∞, g ∈ C 1 (I, R)
satisfies g (t) < t, g 0 (t) ≥ 0, and g (t) → ∞ as t → ∞ and f ∈ C (R, R) such that
f is nondecreasing, xf (x) > 0 for x 6= 0.
We consider only those solutions of Eq. (1.1) which are defined and nontrivial
for all sufficiently large t. Such a solution is called oscillatory if it has arbitrarily
large zeros, otherwise it is called nonoscillatory.
Note that if x is a solution of Eq. (1.1), then −x is a solution of
0 0
r2 (t) (r1 (t) x0 ) + p (t) x0 + q (t) f ∗ (x (g (t))) = 0 ,
where f ∗ (x) = −f (−x) and xf ∗ (x) > 0 for all x 6= 0. Since f ∗ and f are of the
same class, we may restrict our attention only to a positive solution of Eq. (1.1)
whenever a nonoscillatory solution of Eq. (1.1) is concerned.
In recent years, the oscillatory and asymptotic behavior of differential equations
and their applications have been and still are receiving intensive attention. In fact,
there are several monographs and hundreds of research papers for ordinary and
functional differential equations, see for example the monographs Agarwal et al.
[1]–[2], Erbe et al. [8], Gyori and Ladas [10], and Swanson [16].
2000 Mathematics Subject Classification: primary 34K11; secondary 34C10.
Key words and phrases: oscillation, third order, functional differential equation.
Received August 8, 2008. Editor F. Neuman.
2
R. P. AGARWAL, M. F. AKTAS AND A. TIRYAKI
Determining oscillation criteria in particularly for second order differential
equations has received a great deal of attention in the last few years. Compared to
second order differential equations, the study of oscillation and asymptotic behavior
of third order differential equations has received considerably less attention in
the literature. We obtain some new results in this paper are motivated by recent
of [3, 4, 5, 9, 15, 17] and insure that every solution of Eq. (1.1) is oscillatory or
converges to zero. For general interest on oscillation results we refer, for example,
to Erbe [7], Grace et al. [9], Parhi and Das [11], Philos and Sficas [13], Seman [14],
Tiryaki and Yaman [18], and the references cited therein.
In this section we state and prove some lemmas which we will use in the proof
of our main results.
For the sake of brevity, we define
0
L0 x (t) = x (t) ,
Li x (t) = ri (t) (Li−1 x (t)) ,
L3 x (t) = (L2 x (t))
0
for
i = 1, 2 ,
t∈I.
So Eq. (1.1) can be written as
L3 x (t) +
p (t)
L1 x (t) + q (t) f (x (g (t))) = 0 .
r1 (t)
Define the functions
Z
t
R1 (t, s) =
s
Z
t
R12 (t, s) =
s
Z t
du
du
,
R2 (t, s) =
,
and
r1 (u)
r
(u)
2
s
Z τ
1
du
dτ , t0 ≤ s ≤ t < ∞ .
r1 (τ ) s r2 (u)
We assume that
(1.2)
R1 (t, t0 ) → ∞ as t → ∞ ,
(1.3)
R2 (t, t0 ) → ∞ as t → ∞ ,
and
(1.4)
R2 (t, t0 ) < ∞
as t → ∞ .
Moreover we shall assume that the function f satisfies conditions:
(1.5)
(1.6)
−f (−uv) ≥ f (uv) ≥ f (u) f (v)
f (u)
≥ K > 0,
u
K
is a real constant, u 6= 0 ,
and
(1.7)
u
→0
f (u)
for uv > 0 ,
as u → 0 .
OSCILLATION CRITERIA
3
Definition 1. The Eq. (1.1) is called superlinear if the function f for every > 0
satisfies
Z ±∞
du
(1.8)
< ∞,
f
(u)
±
and Eq. (1.1) is called sublinear if f satisfies
Z ±
du
(1.9)
< ∞ for every
f
(u)
0
> 0.
Let us give examples of the functions which satisfy the conditions (1.5) and (1.8)
or (1.9).
α
Example 1. The functions f1 and f2 : R → R, where f1 (u) = |u| sgn u, α > 0
2α
|u| sgn u
and f2 (u) =
6 0
α , α > 0 are continuous on R, satisfy uf (u) > 0 for u =
1 + |u|
and conditions nondecreasing of f and (1.5). Further, function f1 satisfies (1.8) for
α > 1 and (1.9) for 0 < α < 1. The function f2 satisfies (1.8) for α > 1.
Lemma 1. Suppose that
r2 (t)z 0
0
+
p (t)
z=0
r1 (t)
is nonoscillatory. If x is a nonoscillatory solution of (1.1) on [T, ∞), T ≥ t0 , then
there exists a t1 ∈ [ T, ∞) such that either x (t) L1 x (t) > 0 or x (t) L1 x (t) < 0 for
all t ≥ t1 .
The reader can refer to [17, Lemma 1] for the proof of Lemma 1.
Lemma 2. Let ρ2 be a sufficiently smooth positive function defined on [t0, ∞), set
0
φ (t) = r1 (t) (r2 (t) ρ02 (t)) + ρ2 (t) p (t) ,
and (1.6) hold. Suppose that there exists a t1 ≥ T ≥ t0 such that
ρ02 (t) ≥ 0 = ,
Z
φ (t) ≥ 0 ,
∞
(1.10)
(Kρ2 (s) q (s) − φ0 (s)) ds = ∞ ,
t1
where Kρ2 (t) q (t) − φ0 (t) ≥ 0 for all t ∈ [t1 , ∞) and not identically zero in any
subinterval of [t1 , ∞). If (1.2) holds and x be a nonoscillatory solution of Eq. (1.1)
which satisfies x (t) L1 x (t) ≤ 0 for all t ≥ t1 , then lim x (t) = 0.
t→∞
The reader can refer to [4, Lemma 2.4] for the proof of Lemma 2.
Remark 1. When
φ0 (t) ≤ 0
(1.11)
in Lemma 2, we can take
Z
(1.12)
∞
ρ2 (s) q (s) ds = ∞
to replace (1.10). Hence the condition (1.6) fails.
4
R. P. AGARWAL, M. F. AKTAS AND A. TIRYAKI
Lemma 3. Let the assumption (1.3) hold. If x is a nonoscillatory solution of Eq.
(1.1) which satisfies x (t) L1 x (t) ≥ 0 for all large t, then there exists a t1 ≥ t0 such
that
(1.13)
L0 x (t) Lk x (t) > 0 ,
k = 0, 1, 2 ; L0 x (t) L3 x (t) ≤ 0
for all t ≥ t1 .
A nonoscillatory solution x of Eq. (1.1) is said to have property V2 if it satisfies
the inequalities (1.13).
Lemma 4. Let x be a solution of (1.1). If x has property V2 for every large t,
then there exists t1 ≥ T ≥ t0 such that either
(1.14)
x (t) ≥ R12 (t, t1 ) L2 x (t) ,
t ≥ t1
or
(1.15)
L1 x (t) ≥ R2 (t, t1 ) L2 x (t) ,
t ≥ t1
or
(1.16)
x (t) ≥
R12 (t, t1 )
L1 x (t) ,
R2 (t, t1 )
t ≥ t1 .
The reader can refer to [6] for the condition (1.16) and [17, Lemma 2] for the
condition (1.15).
2. Main Results
Theorem 1. Let the hypotheses of Lemmas 1–3 and (1.5), (1.11) hold. If the first
order delay equation
(2.1) y 0 (t) +
p (t)
R2 (g (t) , T ) y (g (t)) + q (t) f (R12 (g (t) , T )) f (y (g (t))) = 0
r1 (t)
for every T ≥ t0 is oscillatory, then every solution x of Eq. (1.1) is either oscillatory
or satisfies limt→∞ x (t) = 0.
Proof. Let x be a nonoscillatory solution of Eq. (1.1) on [T, ∞), T ≥ t0 . Without
loss of generality, we may assume that x (t) > 0 and x (g (t)) > 0 for t ≥ T1 ≥ T .
From Lemma 1 it follows that L1 x (t) > 0 or L1 x (t) < 0 for t ≥ t1 ≥ T1 . If
L1 x (t) > 0 for t ≥ t1 , then x has property V2 for large t from Lemma 3. From
Lemma 4, we obtain (1.14) and (1.15). Now there exists a t2 ≥ t1 such that
x (g (t)) ≥ R12 (g (t) , t1 ) L2 x (g (t))
L1 x (g (t)) ≥ R2 (g (t) , t1 ) L2 x (g (t))
and
for t ≥ t2 .
OSCILLATION CRITERIA
5
From Eq. (1.1), we have
−L3 x (t) =
p (t)
L1 x (t) + q (t) f (x (g (t)))
r1 (t)
≥
p (t)
R2 (g (t) , t1 ) L2 x (g (t)) + q (t) f (R12 (g (t) , t1 ) L2 x (g (t)))
r1 (t)
≥
p (t)
R2 (g (t) , t1 ) L2 x (g (t)) + q (t) f (R12 (g (t) , t1 )) f (L2 x (g (t))) ,
r1 (t)
for t ≥ t2 . Setting y (t) = L2 x (t) > 0 for t ≥ t2 , we obtain
y 0 (t) +
p (t)
R2 (g (t) , t1 ) y (g (t)) + q (t) f (R12 (g (t) , t1 )) f (y (g (t))) ≤ 0
r1 (t)
for t ≥ t2 . Integrating the above inequality from t to u and letting u → ∞, we have
Z ∞
p (s)
y (t) ≥
R2 (g (s) , t1 ) y (g (s))
r
1 (s)
t
+ q (s) f (R12 (g (s) , t1 )) f (y (g (s))) ds .
As in [12], it is easy to conclude that there exists a positive solution y (t) of Eq. (2.1)
with limt→∞ y (t) = 0, which contradictions the fact that Eq. (2.1) is oscillatory.
Let x (t) > 0, L1 x (t) < 0, t ≥ t1 . By Remark 1 we have limt→∞ x (t) = 0. The
proof is complete.
Corollary 1. Let the hypotheses of Lemmas 1–3 hold. If the first order delay
equation
p (t)
(2.2)
y 0 (t) + Kq (t) R12 (g (t) , T ) +
R2 (g (t) , T ) y (g (t)) = 0
r1 (t)
for some K > 0 and every T ≥ t0 is oscillatory, then every solution x of Eq. (1.1)
is either oscillatory or satisfies limt→∞ x (t) = 0.
Theorem 2. Let the hypotheses of Lemmas 1–3 hold. If
Z t p (s)
(2.3)
lim sup
Kq (s) R12 (g (s) , T ) +
R2 (g (s) , T ) ds > 1
r1 (s)
t→∞
g(t)
for some K > 0 and every T ≥ t0 , then every solution x of Eq. (1.1) is either
oscillatory or satisfies limt→∞ x (t) = 0.
Proof. Proceeding as in the proof of Theorem 1, we obtain x has property V2 for
large t. From Lemma 4, we obtain (1.14) and (1.15). Now there exists a t2 ≥ t1
such that
x (g (t)) ≥ R12 (g (t) , t1 ) L2 x (g (t))
L1 x (g (t)) ≥ R2 (g (t) , t1 ) L2 x (g (t))
and
for t ≥ t2 .
6
R. P. AGARWAL, M. F. AKTAS AND A. TIRYAKI
Integrating Eq. (1.1) from g (t) to t, we have
Z t p (s)
−L2 x (t) + L2 x (g (t)) =
L1 x (s) + q (s) f (x (g (s))) ds
r1 (s)
g(t)
Z
t
L2 x (g (t)) ≥
g(t)
t
Z
t
p (s)
L1 x (g (s)) + Kq (s) x (g (s))
r1 (s)
ds
p (s)
R2 (g (s) , t1 ) L2 x (g (s)) + Kq (s) R12 (g (s) , t1 ) L2 x (g (s)) ds
r1 (s)
g(t)
Z t p (s)
≥ L2 x (g (t))
Kq (s) R12 (g (s) , t1 ) +
R2 (g (s) , t1 ) ds .
r1 (s)
g(t)
Z
≥
Hence,
1≥
p (s)
Kq (s) R12 (g (s) , t1 ) +
R2 (g (s) , t1 ) ds for t ≥ t2 .
r1 (s)
g(t)
Taking limsup of both sides of the above inequality as t → ∞, we arrive at a
contraction to condition (2.3).
Let x (t) > 0, L1 x (t) < 0, t ≥ t1 . By Lemma 2 we have limt→∞ x (t) = 0. The
proof is complete.
Example 2. Consider the third order delay equation
1
1
3π
= 0,
(2.4)
x000 (t) + 2 x0 (t) + 1 − 2 x t −
4t
4t
2
t≥
3π
.
2
It is easy to check that all conditions of Theorem 2 are satisfied and hence every
solution x (t) of Eq. (2.4) is either oscillatory or satisfies limt→∞ x (t) = 0. An
example of such a solution is x (t) = sin t.
Theorem 3. Let the hypotheses of Lemmas 1–3 hold. If
Z t 1
p (s)
R2 (g (s) , T ) ds >
(2.5)
lim inf
Kq (s) R12 (g (s) , T ) +
t→∞
r1 (s)
e
g(t)
for some K > 0 and any T ≥ t0 , then every solution x of Eq. (1.1) is either
oscillatory or satisfies limt→∞ x (t) = 0.
Proof. Proceeding as in the proof of Theorem 2, we obtain
−L3 x (t) =
p (t)
L1 x (t) + q (t) f (x (g (t)))
r1 (t)
−L3 x (t) ≥
p (t)
L1 x (t) + Kq (t) x (g (t))
r1 (t)
≥
p (t)
R2 (g (t) , t1 ) L2 x (g (t)) + Kq (t) R12 (g (t) , t1 ) L2 x (g (t)) ,
r1 (t)
OSCILLATION CRITERIA
7
for t ≥ t2 . Setting y (t) = L2 x (t) > 0 for t ≥ t2 , we obtain
p (t)
R2 (g (t) , t1 ) y (g (t)) + Kq (t) R12 (g (t) , t1 ) y (g (t)) ≤ 0
r1 (t)
p (t)
0
R2 (g (t) , t1 ) y (g (t)) ≤ 0
y (t) + Kq (t) R12 (g (t) , t1 ) +
r1 (t)
y 0 (t) +
for t ≥ t2 . By known results, see [2, 10, 12], we arrive at the desired contradiction.
Let x (t) > 0, L1 x (t) < 0, t ≥ t1 . By Lemma 2 we have limt→∞ x (t) = 0. The
proof is complete.
Example 3. Consider the third order equation
1
(2.6)
x000 (t) + e−2t+2 x0 (t) + x (t − 1) 1 + x2 (t − 1) = 0 , t ≥ 1 .
e
It is easy to check that all conditions of Theorem 3 are satisfied and hence every
solution x (t) of Eq. (2.6) is either oscillatory or satisfies limt→∞ x (t) = 0. One
such solution of Eq. (2.6) is x (t) = e−t .
Theorem 4. Let the hypotheses of Lemmas 1–3 and (1.5), (1.7), (1.11) hold. If
Z t
(2.7)
lim sup P (t)
q (s) f (R12 (g (s) , T )) ds > 0 ,
t→∞
where P (t) = 1
.
1−
g(t)
Rt
p(s)
R
g(t) r1 (s) 2
(g (s) , T ) ds ≥ 0 for every t ≥ T ≥ t0 and
not identically zero in any subinterval of [T, ∞), then every solution x of Eq. (1.1)
is either oscillatory or satisfies limt→∞ x (t) = 0.
Proof. Proceeding as in the proof of Theorem 1, we obtain
p (t)
−L3 x (t) =
L1 x (t) + q (t) f (x (g (t)))
r1 (t)
p (t)
R2 (g (t) , t1 ) L2 x (g (t)) + q (t) f (R12 (g (t) , t1 )) f (L2 x (g (t))) ,
≥
r1 (t)
for t ≥ t2 ≥ t1 . Integrating the above inequality from g (t) to t, we have
Z t p(s)
−L2 x(t) + L2 x(g(t)) ≥
R2 (g(s), t1 )L2 x(g(s))
g(t) r1 (s)
+ q(s)f (R12 ((s), t1 ))f (L2 x(g(s))) ds
Z
t
L2 x (g (t)) ≥ L2 x (g (t))
g(t)
Z
p (s)
R2 (g (s) , t1 ) ds + f (L2 x (g (t)))
r1 (s)
t
×
q (s) f (R12 (g (s) , t1 )) ds
g(t)
L2 x (g (t))
≥ P (t)
f (L2 x (g (t)))
Z
t
q (s) f (R12 (g (s) , t1 )) ds ,
g(t)
t ≥ t2 ≥ t1 .
8
R. P. AGARWAL, M. F. AKTAS AND A. TIRYAKI
Taking limsup of both sides of the above inequality as t → ∞, we arrive at a
contraction to condition (2.7).
Let x (t) > 0, L1 x (t) < 0, t ≥ t1 . By Remark 1 we have limt→∞ x (t) = 0. The
proof is complete.
Corollary 2. When Theorem 4 doesn’t have the condition (1.11), we can take
either
Z t p (s)
(2.8)
lim sup
Kq (s) f (R12 (g (s) , T )) +
R2 (g (s) , T ) ds > 1
r1 (s)
t→∞
g(t)
or
t→∞
t
p (s)
R2 (g (s) , T ) ds > 1
K 2 q (s) R12 (g (s) , T ) +
r1 (s)
g(t)
Z
lim sup
or
Z
2
t
lim sup K P (t)
t→∞
q (s) f (R12 (g (s) , T )) ds > 1
g(t)
to replace (2.7).
Example 4. Consider
(2.9)
x000 (t) +
1 0
x (t) + t1−2γ xγ (t − 1) = 0 ,
4t2
t ≥ 1,
where γ is the ratio of two positive odd integers, 0 < γ < 1. By choosing ρ2 (t) = t2γ ,
we see that all conditions of Theorem 4 are satisfied. Then, every solution x (t) of
Eq. (2.9) is either oscillatory or satisfies limt→∞ x (t) = 0.
Now, we consider g (t) ≤ t.
Theorem 5. Let the hypotheses of Lemmas 1–3 and g (t) ≤ t, (1.5), (1.11) hold.
If the second order equation
R12 (g (t) , T )
p (t)
0
y (g (t)) + q (t) f
f (y (g (t))) = 0
(2.10) (r2 (t) y 0 (t)) +
r1 (t)
R2 (g (t) , T )
for every T ≥ t0 is oscillatory, then every solution x of Eq. (1.1) is either oscillatory
or satisfies limt→∞ x (t) = 0.
Proof. Proceeding as in the proof of Theorem 1, we obtain x has property V2 for
large t. From Lemma 4, we obtain (1.16). Now there exists a t2 ≥ t1 such that
x (g (t)) ≥
R12 (g (t) , t1 )
L1 x (g (t))
R2 (g (t) , t1 )
for t ≥ t2 .
OSCILLATION CRITERIA
9
From Eq. (1.1), we have
p (t)
L1 x (t) + q (t) f (x (g (t)))
r1 (t)
p (t)
R12 (g (t) , t1 )
≥
L1 x (g (t)) + q (t) f
L1 x (g (t))
r1 (t)
R2 (g (t) , t1 )
p (t)
R12 (g (t) , t1 )
≥
L1 x (g (t)) + q (t) f
f (L1 x (g (t)))
r1 (t)
R2 (g (t) , t1 )
−L3 x (t) =
and so
p (t)
L1 x (g (t)) + q (t) f
L1 x (t) L3 x (t) +
r1 (t)
R12 (g (t) , t1 )
R2 (g (t) , t1 )
f (L1 x (g (t))) ≤ 0
for every t ≥ t2 ≥ t1 . By Theorem 1 in [14] the Eq. (2.10) is oscillatory if and only
if the inequality
(2.11) p (t)
R12 (g (t) , t1 )
0
0
y (t) (r2 (t) y (t)) +
y (g (t)) + q (t) f
f (y (g (t))) ≤ 0
r1 (t)
R2 (g (t) , t1 )
is oscillatory, too. This is a contradiction, since y = L1 x (t) is a nonoscillatory
solution of (2.11) for large t.
Let x (t) > 0, L1 x (t) < 0, t ≥ t1 . By Remark 1 we have limt→∞ x (t) = 0. The
proof is complete.
Corollary 3. Let the hypotheses of Lemmas 1–3 and g (t) ≤ t hold. If the second
order equation
R12 (g (t) , T )
p (t)
0
0
(r2 (t) y (t)) + Kq (t)
+
y (g (t)) = 0
R2 (g (t) , T )
r1 (t)
for some K > 0 and every T ≥ t0 is oscillatory, then every solution x of Eq. (1.1)
is either oscillatory or satisfies limt→∞ x (t) = 0.
Example 5. Consider
(2.12)
x000 (t) +
p0 0
q0
x (t) + β x (λt) = 0, t ≥ 1 ,
δ
t
t
0 < λ ≤ 1,
1
p0
, q0 > 0, δ ≥ 2, and β < 3 are some constants. Equation z 00 + δ z =
4
t
q0 λt − 1
00
0 is nonoscillatory (see [16, pp. 45]) and also since y (t) + β
y (λt) = 0 is
t
2
p0
q0 λt − 1
y (λt) = 0 is
oscillatory (see [14, Theorem 6]), equation y 00 (t) +
+ β
tδ
t
2
oscillatory by the generalized Sturm comparison theorem (see [14, Theorem 2]). If
we also choose ρ2 (t) = t2 , from Theorem 5, every solution x (t) of Eq. (2.12) is either
1
oscillatory or satisfies x (t) → 0 as t → ∞. If we take δ = 2, β = 3, λ = 1, p0 =
4
25
1
3
3
2
2
and q0 =
, x1 (t) = , x2 (t) = t cos
ln t , and x3 (t) = t sin
ln t are
4
t
2
2
solutions of Euler Eq. (2.12) and all hypotheses of Theorem 5 are satisfied.
where 0 ≤ p0 ≤
10
R. P. AGARWAL, M. F. AKTAS AND A. TIRYAKI
Theorem 6. Let the hypotheses of Lemmas 1–3 and g (t) ≤ t, (1.5), (1.8), (1.11)
hold. If
Z ∞
R12 (g (s) , T )
(2.13)
q (s) R2 (g (s) , T ) f
ds = ∞ for T ≥ t0 ,
R2 (g (s) , T )
T
then every solution x of Eq. (1.1) is either oscillatory or satisfies limt→∞ x (t) = 0.
Proof. Proceeding as in the proof of Theorem 1, we obtain x has property V2 for
large t. Now there exists a t2 ≥ t1 such that
x (g (t)) ≥
R12 (g (t) , t1 )
L1 x (g (t))
R2 (g (t) , t1 )
for t ≥ t2 .
From Eq. (1.1), we have
−
d
p (t)
L2 x (t) =
L1 x (t) + q (t) f (x (g (t)))
dt
r1 (t)
R12 (g (t) , t1 )
≥ q (t) f
L1 x (g (t))
R2 (g (t) , t1 )
R12 (g (t) , t1 )
≥ q (t) f
f (L1 x (g (t))) ,
R2 (g (t) , t1 )
t ≥ t2 .
Then integrating from t to u ≥ t ≥ t2 , we get
Z u
R12 (g (s) , t1 )
f (L1 x (g (s))) ds
L2 x (t) ≥ L2 x (t) − L2 x (u) ≥
q (s) f
R2 (g (s) , t1 )
t
and from this
Z
L2 x (t) ≥
t
∞
q (s) f
R12 (g (s) , t1 )
R2 (g (s) , t1 )
f (L1 x (g (s))) ds for t ≥ t2 .
Setting y (t) = L1 x (t) > 0 for t ≥ t2 , we obtain
Z ∞
R12 (g (s) , t1 )
0
(2.14) r2 (t) y (t) ≥
q (s) f
f (y (g (s))) ds for t ≥ t2 .
R2 (g (s) , t1 )
t
Since g, y, and f are nondecreasing functions and r2 (t) y 0 (t) is nonincreasing, we
get
Z ∞
R12 (g (s) , t1 )
0
r2 (g (t)) y (g (t)) ≥ f (y (g (t)))
q (s) f
ds for t ≥ t2 .
R2 (g (s) , t1 )
t
Multiplying this inequality by g 0 (t) and dividing it by r2 (g (t)) f (y (g (t))) and
then integrating it from t2 to t ≥ t2 , we have
Z ∞
Z t 0
Z t
y (g (s)) g 0 (s)
g 0 (s)
R12 (g (u) , t1 )
ds ≥
q (u) f
du ds
R2 (g (u) , t1 )
t2 f (y (g (s)))
t2 r2 (g (s))
s
OSCILLATION CRITERIA
11
and from this
Z ∞
Z y(g(t))
du
du
≥
f
(u)
f
(u)
y(g(t2 ))
y(g(t2 ))
Z t
Z t
0
g (s)
R12 (g (u) , t1 )
q (u) f
≥
du ds
R2 (g (u) , t1 )
t2 r2 (g (s))
s
Z t
R12 (g (s) , t1 )
=
[R2 (g (s) , t2 ) − R2 (g (t2 ) , t2 )] q (s) f
ds
R2 (g (s) , t1 )
t2
Z
R12 (g (s) , t1 )
1 t
q (s) R2 (g (s) , t2 ) f
ds
≥
2 t3
R2 (g (s) , t1 )
R2 (g (t) , t2 )
for t ≥ t3 . The
2
last inequality contradicts the assumption (2.13) for large t.
Let x (t) > 0, L1 x (t) < 0, t ≥ t1 . By Remark 1 we have limt→∞ x (t) = 0. The
proof is complete.
for t ≥ t3 , where t3 ≥ t2 is such that R2 (g (t2 ) , t2 ) ≤
Example 6. Consider the third order equation
√
√
2α √ 2α
x t sgn x t
2α +
t−1
1 0
000
= 0,
(2.15)
x (t) + 3 x (t) +
√ α
√
2α+1
t
1 + x t t t−1
1
z = 0 is nonoscillatory (see [16, pp.45]). If we
t3
2
choose ρ2 (t) = t , from Theorem 6, then every solution x (t) of Eq. (2.15) is either
oscillatory or satisfies limt→∞ x (t) = 0.
for t ≥ 1, α > 1. Equation z 00 +
Remark 2. Let g (t) ≤ t, (1.3), and (1.8) hold. If
Z ∞
R12 (g (s) , T )
q (s) R2 (g (s) , T ) f
ds = ∞ for T ≥ t0 ,
R2 (g (s) , T )
T
then equation
0
0
(r2 (t) y (t)) + q (t) f
R12 (g (t) , T )
R2 (g (t) , T )
f (y (g (t))) = 0
is oscillatory (see [14, Theorem 4]).
Theorem 7. Let the hypotheses of Lemmas 1–3 and g (t) ≤ t, (1.5), (1.8), (1.11)
hold. Let there exists a nondecreasing function G ∈ C (R, R) such that f (x) =
|x| G (x) for x ∈ R. Then, if
Z ∞
R12 (g (s) , T )
2
q (s) R2 (g (s) , T ) f
R2 (g (s) , T )
T
!
Z ∞
R12 (g (u) , T )
(2.16)
×
q (u) f
du ds = ∞
R2 (g (u) , T )
g(s)
12
R. P. AGARWAL, M. F. AKTAS AND A. TIRYAKI
for T ≥ t0 , and
Z
±∞
dx
< ∞,
G (x)
±
for every ε > 0, then every solution x of Eq. (1.1) is either oscillatory or satisfies
limt→∞ x (t) = 0.
Proof. Proceeding as in the proof of Theorem 6, we obtain x has property V2 for
large t. Then y (t) = L1 x (t) is the nonoscillatory solution of the equation
0
(r2 (t) y 0 (t)) + b (t) G (y (g (t))) = 0 ,
R12 (g (t) , t1 )
where b (t) = q (t) f
y (g (t)) for t ≥ t1 . Then by Remark 2
R2 (g (t) , t1 )
Z ∞
R12 (g (s) , t1 )
y (g (s)) ds < ∞ .
(2.17)
q (s) R2 (g (s) , t1 ) f
R2 (g (s) , t1 )
t1
In the same way as in the proof of Theorem 6 from (2.14) we have
Z ∞
R12 (g (s) , t1 )
0
r2 (t) y (t) ≥ f (y (g (t)))
ds
q (s) f
R2 (g (s) , t1 )
t
Z ∞
R12 (g (s) , t1 )
≥ f (y (g (t2 )))
q (s) f
ds
R2 (g (s) , t1 )
t
for t ≥ t2 . Dividing this inequality by r2 (t) and integrating it from t2 to t ≥ t2 we
get
Z ∞
Z t
1
R12 (g (u) , t1 )
du ds
y (t) ≥ f (L1 x (g (t2 )))
q (u) f
R2 (g (u) , t1 )
t2 r2 (s)
s
Z ∞
Z t
1
R12 (g (u) , t1 )
≥ f (L1 x (g (t2 )))
q (u) f
du ds
R2 (g (u) , t1 )
t2 r2 (s)
t
Z ∞
R12 (g (s) , t1 )
= f (L1 x (g (t2 ))) (R2 (t, t2 ) − R2 (t0 , t2 ))
ds .
q (s) f
R2 (g (s) , t1 )
t
Then there exists a t3 ≥ t2 such that
Z ∞
1
R12 (g (s) , t1 )
y (g (t)) ≥ f (L1 x (g (t2 ))) R2 (g (t) , t2 )
q (s) f
ds
2
R2 (g (s) , t1 )
g(t)
for t ≥ t3 . This inequality and (2.17) contradict the condition (2.16).
Let x (t) > 0, L1 x (t) < 0, t ≥ t1 . By Remark 1 we have limt→∞ x (t) = 0. The
proof is complete.
Example 7. The equation
x000 (t) + t−3 x0 (t) + t−5/2 x3 t1/3 = 0 ,
t ≥ 1,
satisfies the assumptions of Theorem 7 but the condition (2.13) of Theorem 6 does
not hold.
There are many sufficient conditions for the oscillation of equation (2.10) in the
literature. The reader can refer to [1]–[2], [14] for them.
OSCILLATION CRITERIA
13
Theorem 8. Let the hypotheses of Lemmas 1–3 and g (t) ≤ t, (1.5), (1.9), (1.11)
hold. If
Z ∞
(2.18)
q (s) f (R12 (g (s) , T )) ds = ∞ for T ≥ t0 ,
then every solution x of Eq. (1.1) is either oscillatory or satisfies limt→∞ x (t) = 0.
Proof. Proceeding as in the proof of Theorem 1, we obtain x has property V2 for
large t. From Eq. (1.1), we have
−
p (t)
d
L2 x (t) =
L1 x (t) + q (t) f (x (g (t)))
dt
r1 (t)
≥ q (t) f (R12 (g (t) , t1 ) L2 x (g (t)))
≥ q (t) f (R12 (g (t) , t1 )) f (L2 x (t))
or
d
(L2 x (t))
dt
≥ q (t) f (R12 (g (t) , t1 ))
f (L2 x (t))
−
for
t ≥ t2 ≥ t1 .
Integrating the above inequality from t2 to t, we have
Z t
Z L2 x(t2 )
du
≥
q (s) f (R12 (g (s) , t1 )) ds .
f (u)
t2
L2 x(t)
Taking lim of both sides of the above inequality as t → ∞, we obtain at a contraction
to condition (2.18).
Let x (t) > 0, L1 x (t) < 0, t ≥ t1 . By Remark 1 we have limt→∞ x (t) = 0. The
proof is complete.
Example 8. Consider
α
(2.19) x000 (t)+
1 0
25 (λt)
α−1
x (t)+
|x (λt)|
x (λt) = 0, t ≥ 1 ,
2
4t
4 t4
0 < α, λ < 1.
By choosing ρ2 (t) = t2 , it is easy to check that all conditions of Theorem 8 are
satisfied. Then every solution x (t) of Eq. (2.19) is either oscillatory or satisfies
1
limt→∞ x (t) = 0. Observe that x (t) = is a solution of Eq. (2.19).
t
Theorem 9. Let g (t) ≤ t and the function f satisfy the condition
(2.20)
lim inf |f (u)| > 0 .
|u|→∞
If
Z
(2.21)
∞
q (t) dt = ∞ ,
then every solution x of Eq. (1.1) is either oscillatory or satisfies limt→∞ x (t) = 0.
14
R. P. AGARWAL, M. F. AKTAS AND A. TIRYAKI
Proof. Proceeding as in the proof of Theorem 1, we obtain x has property V2 for
large t. Since x has property V2 , limt→∞ x (t) exists. If limt→∞ x (t) = ∞, then
from (2.20) and (2.21) we obtain
Z ∞
(2.22)
q (t) f (x (g (t))) dt = ∞ .
If limt→∞ x (t) = K < ∞, then from (2.21) and the continuity f (2.22) holds, too.
Integrating the inequality L3 x (t) + q (t) f (x (g (t))) ≤ 0 from t1 to t ≥ t1 and using
(2.22) we get L2 x (t) < 0 for all sufficiently large t, a contradiction.
Let x (t) > 0, L1 x (t) < 0, t ≥ t1 . By Remark 1 (ρ2 (t) = 1) we have limt→∞ x (t) =
0. The proof is complete.
Example 9. Consider the third order equation
00
1 0
1
1
1
(2.23)
x (t) + 3 x0 (t) + x (t − ln t) 1 +
= 0,
t
4t
t
1 + x2 (t − ln t)
for t ≥ 1. It is easy to check that all conditions of Theorem 9 are satisfied. Then
every solution x (t) of Eq. (2.23) is either oscillatory or satisfies limt→∞ x (t) = 0.
Now, we consider
R2 (t, t0 ) < ∞ .
(1.4)
Theorem 10. Let the hypotheses of Lemmas 1–2 and (1.4), (1.5), (1.11) hold. In
addition to the first order delay equation
(2.1) y 0 (t) +
p (t)
R2 (g (t) , T ) y (g (t)) + q (t) f (R12 (g (t) , T )) f (y (g (t))) = 0
r1 (t)
for every T ≥ t0 is oscillatory. If
Z ∞
Z u
1
(Dq (s) f (R1 (g (s) , T )) f (R2 (∞, g (s)))
r2 (u) T
T
(2.24)
+
p (s)
R2 (∞, g (s)) ds du = ∞
r1 (s)
for every D > 0 and any T ≥ t0 , then every solution x of Eq. (1.1) is either
oscillatory or satisfies limt→∞ x (t) = 0.
Proof. Let x be a nonoscillatory solution of (1.1) on [T, ∞), T ≥ t0 . Without loss
of generality, we may assume that x (t) > 0 and x (g (t)) > 0 for t ≥ T1 ≥ T . From
Lemma 1 it follows that L1 x (t) > 0 or L1 x (t) < 0 for t ≥ t1 ≥ T1 . There are three
possibility to consider:
(i) L1 x (t) > 0, L2 x (t) > 0, L3 x (t) ≤ 0 for t ≥ t1 ;
(ii) L1 x (t) > 0, L2 x (t) < 0, L3 x (t) ≤ 0 for t ≥ t1 ; and
(iii) L1 x (t) < 0 for t ≥ t1 .
OSCILLATION CRITERIA
15
Case (i): The proof is exactly the same as that Theorem 1 – Case (i).
Case (ii): There exists a t2 ≥ t1 such that
x (t) ≥ R1 (t, t1 ) L1 x (t)
t ≥ t2
for
and so there exists a t3 ≥ t2 such that
(2.25) x (g (t)) ≥ R1 (g (t) , t1 ) L1 x (g (t)) := R1 (g (t) , t1 ) v (g (t))
for t ≥ t3 ,
where v (t) = L1 x (t). Using (2.25) and (1.5) in Eq. (1.1), we find
(2.26)
0
(r2 (t) v 0 (t)) +
p (t)
v (g (t)) + q (t) f (R1 (g (t) , t1 )) f (v (g (t))) ≤ 0
r1 (t)
for t ≥ t3 . Clearly, v (t) > 0 and v 0 (t) < 0 for t ≥ t3 . Now, for s ≥ t ≥ t3 one can
easily see that
(2.27)
− r2 (s) v 0 (s) ≥ −r2 (t) v 0 (t)
s ≥ t ≥ t3 .
for
Dividing (2.27) by r2 (s) and integrating from t to u ≥ t ≥ t3 , we have
v (t) ≥ v (t) − v (u) ≥ −r2 (t) v 0 (t) R2 (u, t) .
Letting u → ∞ in the above inequality, we get
(2.28)
v (t) ≥ −r2 (t) v 0 (t) R2 (∞, t)
for
t ≥ t3 .
Combining (2.28) with the inequality
−r2 (t) v 0 (t) ≥ −r2 (t3 ) v 0 (t3 )
for t ≥ t3 ,
which implied by (2.27), we find
v (t) ≥ −r2 (t3 ) v 0 (t3 ) R2 (∞, t)
for t ≥ t3 .
Thus, there exists a constant b > 0 and a t4 ≥ t3 such that
(2.29)
v(g(t)) ≥ bR2 (∞, g(t))
for t ≥ t4 .
Integrating inequality (2.26) from t3 to t, we have
Z t
p(s)
v(g(s)) + q(s)f (R1 (g(s), t1 ))f (v(g(s))) ds
t3 r1 (s)
≤ r2 (t3 )v 0 (t3 ) − r2 (t)v 0 (t) .
Using Eq. (2.29) and (1.5) in the above inequality, we get
Z t
1
f (b) q (s) f (R1 (g (s) , t1 )) f (R2 (∞, g (s)))
r2 (t) t3
p (s)
+b
R2 (∞, g (s)) ds ≤ −v 0 (t) , t ≥ t4 .
r1 (s)
16
R. P. AGARWAL, M. F. AKTAS AND A. TIRYAKI
Integrating the above inequality from t4 to t, we find
Z t
Z τ
1
b
Dq (s) f (R1 (g (s) , t1 )) f (R2 (∞, g (s)))
t4 r2 (τ ) t3
p (s)
+
R2 (∞, g (s)) ds dτ ≤ v (t4 ) < ∞ ,
r1 (s)
where D =
f (b)
is a constant. This inequality implies
b
Z ∞
Z τ
1
Dq (s) f (R1 (g (s) , t1 )) f (R2 (∞, g (s)))
r2 (τ ) t3
t4
p (s)
+
R2 (∞, g (s)) ds dτ < ∞ ,
r1 (s)
which contradictions condition (2.24).
Case (iii): Let x (t) > 0, L1 x (t) < 0, t ≥ t1 . By Remark 1 we have limt→∞ x (t) = 0.
The proof is complete.
Corollary 4. Let the hypotheses of Lemmas 1–2 and (1.4) hold. In addition to
the first order delay equation
p (t)
0
(2.2)
y (t) + Kq (t) R12 (g (t) , T ) +
R2 (g (t) , T ) y (g (t)) = 0
r1 (t)
for some K > 0 and every T ≥ t0 is oscillatory. If
(2.30)
Z ∞
Z u
1
p (s)
R2 (∞, g (s)) Kq (s) R1 (g (s) , T ) +
ds du = ∞
r2 (u) T
r1 (s)
T
for some K > 0 and any T ≥ t0 , then every solution x of Eq. (1.1) is either
oscillatory or satisfies limt→∞ x (t) = 0.
Theorem 11. Let the hypotheses of Lemmas 1–2 and (1.4) hold. Then every
solution x of Eq. (1.1) is either oscillatory or satisfies limt→∞ x (t) = 0 if one of
the following conditions holds:
(I1 ) Condition (2.30) and
Z t p (s)
(2.4)
lim sup
Kq (s) R12 (g (s) , T ) +
R2 (g (s) , T ) ds > 1
r1 (s)
t→∞
g(t)
for some K > 0 and every T ≥ t0 .
(I2 ) Condition (2.30) and
Z t p (s)
1
(2.6)
lim inf
Kq (s) R12 (g (s) , T ) +
R2 (g (s) , T ) ds >
t→∞
r1 (s)
e
g(t)
for some K > 0 and any T ≥ t0 .
OSCILLATION CRITERIA
17
(I3 ) Conditions (1.5), (1.7), (1.11), (2.24), and
Z t
q (s) f (R12 (g (s) , T )) ds > 0
(2.8)
lim sup P (t)
t→∞
g(t)
for any T ≥ t0 .
(I4 ) Conditions g (t) ≤ t, (1.5), (1.8), (1.11), (2.24), and
Z ∞
R12 (g (s) , T )
q (s) R2 (g (s) , T ) f
ds = ∞
(2.13)
R2 (g (s) , T )
T
for T ≥ t0 .
(I5 ) Conditions g (t) ≤ t, (1.5), (1.9), (1.11), (2.24), and
Z ∞
(2.19)
q (s) f (R12 (g (s) , T )) ds = ∞
for T ≥ t0 .
Remark 3. We note that conditions of theorems can be changed when the
conditions are satisfied both (1.5) and (1.6) at the same time (see Corollary 2).
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Department of Mathematical Sciences
Florida Institute of Technology
Melbourne, FL 32901, USA
E-mail: agarwal@fit.edu
Department of Mathematics, Gazi University
Faculty of Arts and Sciences
Teknik- okullar, 06500 Ankara, Turkey
E-mail: mfahri@gazi.edu.tr
Department of Mathematics and Computer Sciences
Izmir University, Faculty of Arts and Sciences
35350 Uckuyular, Izmir, Turkey
E-mail: aydin.tiryaki@izmir.edu.tr